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### 100 Animals, 100 Dollars, No Algebra

```Date: 09/05/2003 at 12:11:58
From: Melissa
Subject: How Many of Each Animal Can I Get

You have exactly \$100.00 to spend.  You must get 100 animals.  The
chicks cost \$0.10 each.  The pigs cost \$2.00 each.  The sheep cost
\$5.00 each.  You must get some of each animal.  How many of each
animal can you get?

I have found a way to get 100 animals (80 chicks, 4 pigs, 16 sheep) by
spending \$96.00, and a way to get 98 animals (80 chicks, 1 pig, 17
sheep) by spending \$100.00, but I can't find a way to get 100 animals
for \$100.

```

```
Date: 09/05/2003 at 15:16:36
From: Doctor Greenie
Subject: Re: How Many of Each Animal Can I Get

Hi, Melissa --

Here are a couple of links to pages in the Dr. Math archives
containing problems which are similar to yours, in case you want to
try to understand the formal algebraic approach to this type of
problem:

http://mathforum.org/library/drmath/view/53051.html
http://mathforum.org/library/drmath/view/57395.html

But let me see if I can show you an approach that doesn't require
alegbra.

We need to have the numbers of sheep, pigs, and chickens adding to
100; and we need the total cost of the animals, at \$5 per sheep, \$2
per pig, and \$.10 per chicken, adding to \$100.00.

The first thing we can logically conclude about this problem which
will make it easier to find the solution is that the number of
chickens must be a multiple of 10.  The cost of each sheep and of
each pig are whole-dollar amounts, so the total cost of the sheep
and pigs is a whole-dollar amount.  And the total cost (\$100.00) is
a whole-dollar amount -- so the total cost of the chickens must be a
whole-dollar amount.  Since the chickens cost \$.10 each, they must
be bought in groups of 10 to make the total cost a whole-dollar
amount.

So we know the number of chickens can be only one of the following:

10 20 30 40 50 60 70 80 90

Now let's look at these different possibilities for the number of
chickens; for each of the possibilities, we will see if we can find
numbers of sheep and pigs so that the total number of animals is 100
and the total cost is \$100.  We can arrange the information in a
table.  For each possible number of chickens, we will show (a) the
cost of that many chickens; (b) the remaining number of animals; and
(c) the range of cost of that number of remaining animals, knowing
that the pigs (the cheaper) are \$2 each and the sheep (the more
expensive) are \$5 each.

chickens            sheep &amp; pigs
number  cost        number      cost      total cost
-------------------------------------------------------
10     \$1         90   \$180 to \$450   \$181 to \$451
20     \$2         80   \$160 to \$400   \$162 to \$402
30     \$3         70   \$140 to \$350   \$143 to \$353
40     \$4         60   \$120 to \$300   \$124 to \$304
50     \$5         50   \$110 to \$250   \$115 to \$255
60     \$6         40   \$ 80 to \$200   \$ 86 to \$206
70     \$7         30   \$ 60 to \$150   \$ 67 to \$157
80     \$8         20   \$ 40 to \$100   \$ 42 to \$108
90     \$9         10   \$ 20 to \$ 50   \$ 29 to \$ 59

From this table, we see that we must have at least 60 chickens -- if
we have 50 chickens or fewer, the cost of the remaining animals is
more than the \$100 we have to spend, and so the total cost is more
than that \$100.  We can also see that we can't have 90 chickens,
because the most we can spend on the other 10 animals is \$50, and
that combination use up only \$59 of our \$100.  The number of
chickens must be either 60, 70, or 80 -- because those are the only
numbers of chickens which allow us to spend a total of \$100 on a
total of 100 animals.

From here, the solution using algebra is quite simple, and finishing
the problem without algebra is quite tedious.  But let's do it....

For each of the remaining possible numbers of chickens, let's do the
following:

(1) look only at combinations of sheep and pigs which make the total
number of animals 100

(2) look first at how much we spend in all if all the remaining
animals are pigs (the cheaper of the two)

(3) figure out how much more we spend each time we "trade in" a pig
for a sheep

(4) from (2) and (3), determine if we can spend a total of \$100 with
all the animals

First case: number of chickens = 60

The 60 chickens cost \$6; the remaining 40 animals must cost \$94. If
all the remaining 40 animals are pigs, we spend an additional \$80,
for a total of \$86.  A sheep costs \$3 more than a pig -- so each
time we "trade in" a pig for a sheep, we spend \$3 more.  So the
total dollar amounts we can spend on 100 animals if we have 60
chickens are the following:

pigs  sheep  total cost
------------------------
40      0      86
39      1      89
38      2      92
37      3      95
36      4      98
35      5     101
34      6     104
...

We see that with 60 chickens we can't spend \$100 on 100 animals....

Second case: number of chickens = 70

The 70 chickens cost \$7; the remaining 30 animals must cost \$93. If
all the remaining 30 animals are pigs, we spend an additional \$60,
for a total of \$67.  A sheep costs \$3 more than a pig -- so each
time we "trade in" a pig for a sheep, we spend \$3 more.  So the
total dollar amounts we can spend on 100 animals if we have 70
chickens are the following:

pigs  sheep  total cost
------------------------
30      0      67
29      1      70
28      2      73
27      3      76
26      4      79
25      5      82
24      6      85
23      7      88
22      8      91
21      9      94
20     10      97
19     11     100
...

We CAN spend \$100 with 70 chickens....

The answer to the problem (or at least AN answer) is 70 chickens, 19
pigs, and 11 sheep.

Third case: number of chickens = 80

We have found A solution to the problem.  You can make a table like
the ones above if you want to for the case with 80 chickens.  If you
do, you will find that the total dollar amounts we can spend on 100
animals if we have 80 chickens are the following:

48, 51, 54, 57, ..., 90, 93, 96, 99, 102, 105, and 108

And again we see that we can't spend \$100 on 100 animals if there
are 80 chickens.  So the one solution we found is the only solution.

I hope all this helps.  These problems are a lot easier to solve
when you have the tools of algebra to help you.  But as you see, we
can solve them without algebra.

Please write back if you have any further questions about any of
this.

- Doctor Greenie, The Math Forum
http://mathforum.org/dr.math/
```
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